Math Problem Solver

The system of equations has one solution.\[x = \frac{50}{131}\] \[y = \frac{45}{131}\] \[z = \frac{36}{131}\]

Solve by substitution:

Solve \(x = \frac{x}{10} + \frac{y}{5} + z\) for \(x\). \[x = \frac{2 y}{9} + \frac{10 z}{9}\]

Substitute \(\frac{2 y}{9} + \frac{10 z}{9}\) for \(x\) in each of the remaining equations and simplify. $$\begin{aligned}y &= \frac{9 x}{10} \\ y &= \frac{9 \left(\frac{2 y}{9} + \frac{10 z}{9}\right)}{10} \\ 4 y &= 5 z \end{aligned}$$$$\begin{aligned}x + y + z &= 1 \\ \left(\frac{2 y}{9} + \frac{10 z}{9}\right) + y + z &= 1 \\ 11 y + 19 z &= 9 \end{aligned}$$

Solve \(4 y = 5 z\) for \(y\). \[y = \frac{5 z}{4}\]

Substitute \(\frac{5 z}{4}\) for \(y\) in \(11 y + 19 z = 9\) and simplify. $$\begin{aligned}11 y + 19 z &= 9 \\ 11 \left(\frac{5 z}{4}\right) + 19 z &= 9 \\ z &= \frac{36}{131} \end{aligned}$$

Use substitution of the numerical value of \(z\) to get the values of \(x\) and \( y\). $$\begin{aligned}y &= \frac{5 z}{4} \\ y &= 5 \cdot 36 \cdot \frac{1}{131} \cdot \frac{1}{4} \\ y &= \frac{45}{131}\end{aligned}$$$$\begin{aligned}x &= \frac{2 y}{9} + \frac{10 z}{9} \\ x &= 2 \cdot 45 \cdot \frac{1}{131} \cdot \frac{1}{9} + 10 \cdot 36 \cdot \frac{1}{131} \cdot \frac{1}{9} \\ x &= \frac{50}{131}\end{aligned}$$

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}\frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\\- \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\1 & 1 & 1 &\bigm |& 1\end{bmatrix}$$

Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:

  1. Swap the positions of any two rows.

  2. Multiply any row by a nonzero scalar.

  3. Multiply a row by a nonzero scalar and add it to any other row.

First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\- \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\\frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\end{bmatrix}$$

The leading term of row \(1\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\- \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\\frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\end{bmatrix}$$

Multiply row \(1\) by scalar \(\frac{9}{10}\) and add it to row \(2\).

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & \frac{19}{10} & \frac{9}{10} &\bigm |& \frac{9}{10}\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\\frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\end{bmatrix}$$

Multiply row \(1\) by scalar \(- \frac{9}{10}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & \frac{19}{10} & \frac{9}{10} &\bigm |& \frac{9}{10}\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & - \frac{11}{10} & - \frac{19}{10} &\bigm |& - \frac{9}{10}\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & \frac{19}{10} & \frac{9}{10} &\bigm |& \frac{9}{10}\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & - \frac{11}{10} & - \frac{19}{10} &\bigm |& - \frac{9}{10}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{10}{19}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & 1 & \frac{9}{19} &\bigm |& \frac{9}{19}\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & - \frac{11}{10} & - \frac{19}{10} &\bigm |& - \frac{9}{10}\end{bmatrix}$$

Multiply row \(2\) by scalar \(-1\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & \frac{10}{19} &\bigm |& \frac{10}{19}\\0 & 1 & \frac{9}{19} &\bigm |& \frac{9}{19}\\0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & - \frac{11}{10} & - \frac{19}{10} &\bigm |& - \frac{9}{10}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{4}{5}\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & \frac{10}{19} &\bigm |& \frac{10}{19}\\0 & 1 & \frac{9}{19} &\bigm |& \frac{9}{19}\\0 & 0 & \frac{131}{95} &\bigm |& \frac{36}{95}\\0 & - \frac{11}{10} & - \frac{19}{10} &\bigm |& - \frac{9}{10}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{11}{10}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & \frac{10}{19} &\bigm |& \frac{10}{19}\\0 & 1 & \frac{9}{19} &\bigm |& \frac{9}{19}\\0 & 0 & \frac{131}{95} &\bigm |& \frac{36}{95}\\0 & 0 & - \frac{131}{95} &\bigm |& - \frac{36}{95}\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 0 & \frac{10}{19} &\bigm |& \frac{10}{19}\\0 & 1 & \frac{9}{19} &\bigm |& \frac{9}{19}\\0 & 0 & \frac{131}{95} &\bigm |& \frac{36}{95}\\0 & 0 & - \frac{131}{95} &\bigm |& - \frac{36}{95}\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{95}{131}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & \frac{10}{19} &\bigm |& \frac{10}{19}\\0 & 1 & \frac{9}{19} &\bigm |& \frac{9}{19}\\0 & 0 & 1 &\bigm |& \frac{36}{131}\\0 & 0 & - \frac{131}{95} &\bigm |& - \frac{36}{95}\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{10}{19}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& \frac{50}{131}\\0 & 1 & \frac{9}{19} &\bigm |& \frac{9}{19}\\0 & 0 & 1 &\bigm |& \frac{36}{131}\\0 & 0 & - \frac{131}{95} &\bigm |& - \frac{36}{95}\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{9}{19}\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& \frac{50}{131}\\0 & 1 & 0 &\bigm |& \frac{45}{131}\\0 & 0 & 1 &\bigm |& \frac{36}{131}\\0 & 0 & - \frac{131}{95} &\bigm |& - \frac{36}{95}\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{131}{95}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& \frac{50}{131}\\0 & 1 & 0 &\bigm |& \frac{45}{131}\\0 & 0 & 1 &\bigm |& \frac{36}{131}\\0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot x+ 0 \cdot y+ 0 \cdot z = \frac{50}{131} \\ x = \frac{50}{131}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 1 \cdot y+ 0 \cdot z = \frac{45}{131} \\ y = \frac{45}{131}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 0 \cdot y+ 1 \cdot z = \frac{36}{131} \\ z = \frac{36}{131}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 0 \cdot y+ 0 \cdot z = 0 \\ None = 0\end{aligned}$$